72. The limit of $x^n$

1. If $\varphi (n)$ is positive and $\varphi (n+1)>K\varphi (n)$, where $K>1$, for all values of $n$, then $\varphi (n)\to +\mathrm{\infty }$. These examples are particularly important and several of them will be made use of later in the text. They should therefore be studied very carefully.

[For $$\varphi (n)>K\varphi (n–1)>K^2\varphi (n–2)\cdots >K^{n-1}\varphi (1),$$ from which the conclusion follows at once, as $K^n\to \mathrm{\infty }$.]

2. The same result is true if the conditions above stated are satisfied only when $n\ge n_0$.

3. If $\varphi (n)$ is positive and $\varphi (n+1)<K\varphi (n)$, where $0<K<1$, then $lim\varphi (n)=0$. This result also is true if the conditions are satisfied only when $n\ge n_0$.

4. If $|\varphi (n+1)|<K|\varphi (n)|$ when $n\ge n_0$, and $0<K<1$, then $lim\varphi (n)=0$.

5. If $\varphi (n)$ is positive and $lim\{\varphi (n+1)\}/\{\varphi (n)\}=l>1$, then $\varphi (n)\to +\mathrm{\infty }$.

[For we can determine $n_0$ so that $\{\varphi (n+1)\}/\{\varphi (n)\}>K>1$ when $n\ge n_0$: we may, e.g., take $K$ between $1$ and $l$. Now apply Ex. 1.]

6. If $lim\{\varphi (n+1)\}/\{\varphi (n)\}=l$, where $l$ is numerically less than unity, then $lim\varphi (n)=0$. [This follows from Ex. 4 as Ex. 5 follows from Ex. 1.]

7. Determine the behaviour, as $n\to \mathrm{\infty }$, of $\varphi (n)=n^r{x}^n$, where $r$ is any positive integer.

[If $x=0$ then $\varphi (n)=0$ for all values of $n$, and $\varphi (n)\to 0$. In all other cases $$\frac{\varphi (n+1)}{\varphi (n)}={\left(\frac{n+1}{n}\right)}^{r}x\to x.$$ First suppose $x$ positive. Then $\varphi (n)\to +\mathrm{\infty }$ if $x>1$ (Ex. 5) and $\varphi (n)\to 0$ if $x<1$ (Ex. 6). If $x=1$, then $\varphi (n)=n^r\to +\mathrm{\infty }$. Next suppose $x$ negative. Then $|\varphi (n)|=n^r|x{|}^n$ tends to $+\mathrm{\infty }$ if $|x|\ge 1$ and to $0$ if $|x|<1$. Hence $\varphi (n)$ oscillates infinitely if $x\le -1$ and $\varphi (n)\to 0$ if $-1<x<0$.]

8. Discuss $n^{-r}{x}^n$ in the same way. [The results are the same, except that $\varphi (n)\to 0$ when $x=1$ or $-1$.]

9. Draw up a table to show how $n^k{x}^n$ behaves as $n\to \mathrm{\infty }$, for all real values of $x$, and all positive and negative integral values of $k$.

[The reader will observe that the value of $k$ is immaterial except in the special cases when $x=1$ or $-1$. Since $lim\{(n+1)/n{\}}^k=1$, whether $k$ be positive or negative, the limit of the ratio $\varphi (n+1)/\varphi (n)$ depends only on $x$, and the behaviour of $\varphi (n)$ is in general dominated by the factor $x^n$. The factor $n^k$ only asserts itself when $x$ is numerically equal to $1$.]

10. Prove that if $x$ is positive then $\sqrt[n]{x}\to 1$ as $n\to \mathrm{\infty }$. [Suppose, , $x>1$. Then $x$, $\sqrt{x}$, $\sqrt[3]{x}$, … is a decreasing sequence, and $\sqrt[n]{x}>1$ for all values of $n$. Thus $\sqrt[n]{x}\to l$, where $l\ge 1$. But if $l>1$ we can find values of $n$, as large as we please, for which $\sqrt[n]{x}>l$ or $x>l^n$; and, since $l^n\to +\mathrm{\infty }$ as $n\to \mathrm{\infty }$, this is impossible.]

11. $\sqrt[n]{n}\to 1$. [For $\sqrt[n+1]{n+1}<\sqrt[n]{n}$ if $(n+1{)}^n<n^{n+1}$ or $\{1+(1/n){\}}^n<n$, which is certainly satisfied if $n\ge 3$ (see § 73 for a proof). Thus $\sqrt[n]{n}$ decreases as $n$ increases from $3$ onwards, and, as it is always greater than unity, it tends to a limit which is greater than or equal to unity. But if $\sqrt[n]{n}\to l$, where $l>1$, then $n>l^n$, which is certainly untrue for sufficiently large values of $n$, since $l^n/n\to +\mathrm{\infty }$ with $n$ (Exs. 7, 8).]

12. $\sqrt[n]{n!}\to +\mathrm{\infty }$. [However large $\mathrm{\Delta }$ may be, $n!>{\mathrm{\Delta }}^n$ if $n$ is large enough. For if $u_n={\mathrm{\Delta }}^n/n!$ then $u_{n+1}/u_n=\mathrm{\Delta }/(n+1)$, which tends to zero as $n\to \mathrm{\infty }$, so that $u_n$ does the same (Ex. 6).]

13. Show that if $-1<x<1$ then $$u_n=\frac{m(m–1)\dots (m–n+1)}{n!}{x}^n=(\genfrac{}{}{0ex}{}{m}{n})x^n$$ tends to zero as $n\to \mathrm{\infty }$.

[If $m$ is a positive integer, $u_n=0$ for $n>m$. Otherwise $$\frac{u_{n+1}}{u_n}=\frac{m–n}{n+1}x\to -x,$$ unless $x=0$.]